Question

Find the value of ${\cot }^{-1}1-{\tan }^{-1}2+{\cot }^{-1}3.$

Answer 1

${\cot }^{-1}1-{\tan }^{-1}2+{\cot }^{-1}3={\cot }^{-1}\left(\frac{3-1}{1+3}\right)-{\tan }^{-1}2={\cot }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}2={\cot }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(2\right)={\tan }^1\left(2\right)-{\tan }^{-1}\left(2\right)=0(\because {\tan }^{-1}x={\cot }^{-1}\frac{1}{x},\text{if}x>0)\Rightarrow {\cot }^{-1}1+{\tan }^{-1}2+{\cot }^{-1}3=0$